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Question: (a) An x-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45....

(a) An x-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45.what is the maximum energy of a photon in the radiation?
b) From your answer to (a).guess what order of accelerating voltage (or electrons) is required in such a tube?

Explanation

Solution

An x-ray tube which converts the input electrical power to x-ray. X-ray are types of radiation which are called electromagnetic waves. X-ray is penetrating from high energy electromagnetic radiation. X-ray imaging creates pictures of the inside of your body. The images show the internal bones are parts of our body in black and white shades.

Complete step-by-step solution:
An X-ray tube is an electrical device containing two principle elements cathode and anode. When an electric current is supplied to a tube an electric current flows from cathode to anode , then the electrons lose their energy which results in the generation of X- radiation. In anode X-radiation is produced the quality of X-ray produced can be controlled by adjusting the electrical quantities.
X-ray tube is an energy converter which converts electrical energy into two other forms that is X-radiation and heat.
Energy of photon is inversely proportional to wavelength of light and is given by the equation
E=hcλE=\dfrac{hc}{\lambda } (1)\cdots \cdots (1)
Where, hh = Planck’s constant
cc = speed of light
λ\lambda = wavelength
X –ray has wavelength which ranges from 108{{10}^{-8}} to 1012{{10}^{-12}} meters
From the data we known the value of short wavelength(λ\lambda)=0.45×1010m=45×1012m0.45\times{{10}^{-10}}m=45\times {{10}^{-12}}m
c=3×108c=3\times {{10}^{8}} And h=6.62×1034h=6.62\times {{10}^{-34}}
On Substituting this values in equation (1)
E=6.62×1034×3×10845×1012 E=0.44×1014J E=0.44×10141.6×1019 E=0.275×105eV \begin{aligned} & E=\dfrac{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}{45\times {{10}^{-12}}} \\\ & E=0.44\times {{10}^{-14}}J \\\ & E=\dfrac{0.44\times {{10}^{-14}}}{1.6\times {{10}^{-19}}} \\\ & E=0.275\times {{10}^{5}}eV \\\ \end{aligned}
E=27.6×103eVE=27.6\times {{10}^{3}}eV
The maximum energy of photon is 27.6keV27.6keV
Maximum energy is equal to the peak accelerating potential of the X-ray tube.
Accelerating voltage is responsible for the production of x-ray by providing energy to the photons in order to get an X-ray of 27.6KeV27.6KeV the electron has to possess a minimum27.6KeV27.6KeV of kinetic energy.
So an accelerating voltage of 30KeV30KeV is necessary or required to produce x-rays.

Note: Students this problem is solved by using the formula of energy of photons and photons which do not have mass(rest).X-rays are produced by stopping high energy electrons in a tungsten target. Maximum energy is equal to peak accelerating potential of the x-ray tube and x-rays are scattered within a material.